Chapter 13.5, Problem 56E

a.

To determine

Test whether the data suggests a linear relationship between specific gravity and at least one of the predictors at 1% level of significance.

Answer to Problem 56E

There is sufficient evidence to conclude that the there is a use of linear relationship between specific gravity and at least one of the five predictors number of fibers in springwood, number of fibers in summerwood, percentage of springwood, light absorption in springwood and light absorption in summerwood at 1% level of significance.

Explanation of Solution

Given info:

A sample of 20 mature woods were taken and the number of fibers in springwood, number of fibers in summerwood, percentage of springwood, light absorption in springwood and light absorption in summerwood were noted .

The coefficient of determination $R^2$ is 0.769.

Calculation:

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_1={\beta }_2={\beta }_3={\beta }_4={\beta }_5=0$

That is, there is no use of linear relationship between specific gravity and the five predictors.

Alternative hypothesis:

$H_a:\text{At least one of the }\beta \text{'}\text{s}\ne 0$

That is, there is a use of linear relationship between specific gravity and at least one of the five predictors.

Test statistic:

$f=\frac{\frac{R^2}{k}}{\frac{\left(1-R^2\right)}{\left[n-\left(k+1\right)\right]}}$

Substitute $R^2$ as 0.769, n as 20 and k as 5 because there are five predictors,number of fibers in springwood, number of fibers in summerwood, percentage of springwood, light absorption in springwood and light absorption in summerwood.

$\begin{array}{c}f=\frac{\frac{0.769}{5}}{\frac{\left(1-0.769\right)}{\left[20-\left(5+1\right)\right]}}\\ =\frac{0.154}{\frac{0.231}{14}}\\ =\frac{2.156}{0.231}\\ =9.33\end{array}$

P-value:

Software procedure:

• Click on Graph, select View Probability and click OK.
• Select F, enter 5 in numerator df and 14 in denominator df.
• Under Shaded Area Tab select X value under Define Shaded Area By and select Right tail.
• Choose X value as 9.33.
• Click OK.

Output obtained from MINITAB is given below:

Conclusion:

The P-value is 0.000 and the level of significance is 0.01.

The P-value is lesser than the level of significance.

That is $0.000\left(=P\text{-value}\right)<0.01\left(=\alpha \right)$.

Thus, the null hypothesis is rejected.

Hence, there is sufficient evidence to conclude that there is ause of linear relationship between specific gravity and at least one of the five predictors at 1% level of significance.

b.

To determine

Calculate the adjusted $R_a^2$ value for the full model and when $x_2$ is dropped.

Answer to Problem 56E

The adjusted $R_a^2$ value for the full model is 0.687.

The adjusted $R_a^2$ value for the model when $x_2$ is dropped is 0.707.

Explanation of Solution

Given info:

The $R^2$ value for the model when $x_2$ is dropped is 0.769.

Calculation:

Adjusted $R_a^2$:

$R_a^2=\frac{N\left(R^2\right)-k}{N-k}$

Adjusted $R_a^2$for the full model:

Substitute n as 20,k as 5, $R^2$ value for the full model is 0.769

$\begin{array}{c}{R}_a^2=\frac{\left(N-1\right)R^2-k}{N-k-1}\\ =\frac{\left(20-1\right)\left(0.769\right)-5}{20-5-1}\\ =\frac{\left(19\right)\left(0.769\right)-5}{14}\\ =\frac{9.611}{14}\end{array}$

$\approx 0.687$

Thus, the adjusted $R_a^2$ value for the full model is 0.687.

Adjusted $R_a^2$for the model when $x_2$ is dropped:

Substitute n as 20, k as 4,$R^2$ value for the model when $x_2$ is droppedis 0.769.

$\begin{array}{c}{R}_a^2=\frac{\left(N-1\right)R^2-k}{N-k-1}\\ =\frac{\left(20-1\right)\left(0.769\right)-4}{20-4-1}\\ =\frac{\left(19\right)\left(0.769\right)-4}{15}\\ =\frac{10.611}{15}\end{array}$

$\approx 0.707$

Thus, the adjusted $R_a^2$ value for the model when $x_2$ is droppedis 0.707.

c.

To determine

Identify whether the data suggests that variables $x_1,x_2,x_4$ have zero coefficients.

Test the hypothesis to see whether the variables $x_1,x_2,x_4$ have zero coefficients at 5% level of significance.

Answer to Problem 56E

Yes, the data suggests that variables $x_1,x_2,x_4$ have zero coefficients.

There issufficient evidence to conclude the variables $x_1,x_2,x_4$ have zero coefficients at 5% level of significance.

Explanation of Solution

Given info:

The $R^2$ value after dropping the variables $x_1,x_2,x_4$ is 0.654 and the corresponding SST is 0.0196610.

Calculation:

After dropping the three variables $x_1,x_2,x_4$ there would be $x_3,x_5$ in the model.

The test hypotheses are given below:

Null hypothesis:

$H_0:{\beta }_3={\beta }_5=0$.

That is, there is no use of linear relationship betweenspecific gravity and at least one of the predictors, percentage of springwood and light absorption in summerwood.

Alternative hypothesis:

$H_a:\text{At least }\beta \text{'}\text{s}\ne 0$

That is, there is use of linear relationship between specific gravity and at least one of the predictors, percentage of springwood and light absorption in summerwood.

From the $R^2$ value given in part (a), the sum of squares due to error for the full model ${\text{SSE}}_k$ is calculated as follows:

$\begin{array}{c}{\text{SSE}}_k=\left(1-R^2\right)\text{SST}\\ \text{=}\left(1-0.769\right)0.0196610\\ =\left(0.231\right)\left(0.0196610\right)\\ =0.004542\end{array}$

Similarly, the sum of squares due to error for the reduced model ${\text{SSE}}_l$ is calculated as follows:

$\begin{array}{c}{\text{SSE}}_l=\left(1-R^2\right)\text{SST}\\ \text{=}\left(1-0.654\right)0.0196610\\ =\left(0.346\right)\left(0.0196610\right)\\ =0.006803\end{array}$

Test statistic:

$f=\frac{\frac{\left(SS{E}_l-SS{E}_k\right)}{k-l}}{\frac{SS{E}_k}{\left[n-\left(k+1\right)\right]}}$

Where,

$SS{E}_k$ represents the sum of squares due to error obtained from the full model.

$SS{E}_l$ represents the sum of squares due to error obtained from the reduced model.

n represents the total number of observations.

k represents the number of predictors on the full model.

l represents the number of predictors on the reduced model.

Substitute 0.004542for ${\text{SSE}}_k$, 0.006803 for ${\text{SSE}}_l$, 20 for n, 5 for k and 2 for l .

$\begin{array}{c}f=\frac{\frac{\left(0.006803-0.004542\right)}{5-2}}{\frac{0.004542}{\left[20-\left(5+1\right)\right]}}\\ =\frac{\frac{0.002261}{3}}{\frac{0.004542}{14}}\\ =\frac{0.0007536667}{0.00032443}\\ =2.32\end{array}$

Critical value:

Software procedure:

• Click on Graph, select View Probability and click OK.
• Select F, enter 3 in numerator df and 14 in denominator df.
• Under Shaded Area Tab select Probability under Define Shaded Area By and select Right tail.
• Choose X Value as 2.32.
• Click OK.

Output obtained from MINITAB is given below:

Conclusion:

The P-value is 0.1197 and the level of significance is 0.05.

The P-value is lesser than the level of significance.

That is, $0.1197\left(=P\text{-value}\right)>0.05\left(=\alpha \right)$.

Thus, the null hypothesis is not rejected.

Hence, there is no sufficient evidence to conclude that there is a use of linear relationship betweenspecific gravity and at least one of the predictor percentage of springwood and light absorption in summerwood at 5% level of significance.

Thus, the variables $x_1,x_2,x_4$ can be removed from the model.

d.

To determine

Predict the value of specific gravity when the percentage of springwood is 50 and percentage of light absorption in summerwood is 90.

Answer to Problem 56E

The estimated value for specific gravity when the percentage of springwood is 50 and percentage of light absorption in summerwood is 90 is 0.5386.

Explanation of Solution

Given info:

The mean and standard deviation for the variable $x_3$ is 52.540 and 5.4447 whereas the mean and standard deviation for the variable $x_5$ is 89.195 and 3.6660, respectively.

The estimated regression equation after standardization is

$y=0.5255-0.0236{x^{\prime }}_3+0.0097{x^{\prime }}_5$

Calculation:

The standardized values $x^{\prime }$ is obtained by using the formula:

$x^{\prime }=\frac{x-\overline{x}}{\sigma }$

Where,

$\overline{x}$ represents the mean.

$\sigma$ represents the standard deviation.

The standardized value when mean and standard deviation for the variable $x_3$ is 52.540 and 5.4447 is calculated as follows:

$\begin{array}{c}{x^{\prime }}_3=\frac{x_3-\overline{x_3}}{{\sigma }_3}\\ =\frac{50-52.540}{5.4447}\\ =\frac{-2.540}{5.4447}\\ =-0.467\end{array}$

Thus, the value of ${x^{\prime }}_3$ is –0.467.

The standardized value when the mean and standard deviation for the variable $x_5$ is 89.195 and 3.6660 is calculated as follows:

$\begin{array}{c}{x^{\prime }}_5=\frac{x_5-\overline{x_5}}{{\sigma }_5}\\ =\frac{90-89.195}{3.6660}\\ =\frac{0.805}{3.6660}\\ =0.22\end{array}$

Thus, the value of ${x^{\prime }}_5$ is 0.22.

The estimated value for specific gravity is,

$\begin{array}{c}y=0.5255-0.0236\left(-0.467\right)+0.0097\left(0.22\right)\\ =0.5255+0.0110+0.0021\\ =0.5386\end{array}$

Thus, the estimated value for specific gravity when the percentage of springwood is 50 and percentage of light absorption in summerwood is 90 is 0.5386.

e.

To determine

Find the 95% confidence interval for the estimated coefficient of ${x^{\prime }}_3$.

Answer to Problem 56E

The 95% confidence interval for the estimated coefficient of ${x^{\prime }}_3$ is (–0.0333, –0.0139).

Explanation of Solution

Calculation:

95% confidence interval:

The confidence interval is calculated using the formula:

$\widehat{{\beta }_i}\pm t_{\frac{\alpha }{2},n-\left(k+1\right)}{s}_{{\widehat{\beta }}_i}$

Where,

$\widehat{{\beta }_i}$ is the estimated slope coefficient.

$\alpha$ is the level of significance.

n is the total number of observations.

k is the total number of predictors in the model.

$s_{{\widehat{\beta }}_i}$ is the standard error while calculating the estimated slope coefficient.

Critical value:

Software procedure:

Step-by-step procedure to find the critical value is given below:

• Click on Graph, select View Probability and click OK.
• Select t, enter 17 as Degrees of freedom, in Shaded Area Tab select Probability under Define Shaded Area By and choose Both tails.
• Enter Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

The 95% confidence interval is given below:

$\begin{array}{c}\widehat{{\beta }_i}\pm t_{\frac{\alpha }{2},n-\left(k+1\right)}{s}_{{\widehat{\beta }}_i}=-0.0236\pm t_{\frac{0.05}{2},20-\left(2+1\right)}\left(0.0046\right)\\ =-0.0236\pm t_{0.025,17}\left(0.0046\right)\\ =-0.0236\pm \left(2.110\right)\left(0.0046\right)\\ =-0.0236\pm 0.0097\end{array}$

$=-0.0333,-0.0139$

Thus, the 95% confidence interval for the estimated coefficient of ${x^{\prime }}_3$ is $\left(-0.0333,-0.0139\right)$.

f.

To determine

Find the estimated coefficient and estimated standard deviation of $x_3$ in the unstandardized model.

Answer to Problem 56E

The estimated coefficient of $x_3$ in the unstandardized model is –0.004334.

The estimated standard deviation of ${\widehat{\beta }}_3$ in the unstandardized model is 0.000845.

Explanation of Solution

Given info:

Use the information given in part (d) and (e).

Calculation:

The estimated regression equation for standardized model is,

$y=0.5255-0.0236\left(\frac{x_3-52.540}{5.4447}\right)+0.0097\left(\frac{x_5-89.195}{3.6660}\right)$

The estimated coefficient of $x_3$ in the unstandardized model is calculated as follows:

$\begin{array}{c}{\widehat{\beta }}_3=\frac{-0.0236}{5.4447}\\ =-0.004334\end{array}$

Thus, the estimated coefficient of $x_3$ in the unstandardized model is –0.004334.

The estimate for ${\widehat{\beta }}_3$ is computed using the formula:

${\widehat{\beta }}_3=\frac{{\widehat{\beta }}_3^{*}}{s_x^2}$

The estimated standard deviation for ${\widehat{\beta }}_3$ is the standard deviation of ${\widehat{\beta }}_3^{*}$ multiplied by $\frac{1}{s_x^2}$ and it is given below:

$\begin{array}{c}s\left({\widehat{\beta }}_3\right)=\left(0.0046\right)\frac{1}{5.4447}\\ =0.000845\end{array}$

Thus, the estimated standard deviation of ${\widehat{\beta }}_3$ in the unstandardized model is 0.000845.

g.

To determine

Find the 95% prediction interval for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9.

Answer to Problem 56E

The 95% prediction interval for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9 is(0.489, 0.575).

Explanation of Solution

Given info:

The estimated standard deviation for the model with two predictors is 0.02001. The estimated standard deviation for the predicated value when the coefficients ${x^{\prime }}_{{}_3}$ and ${x^{\prime }}_5$ takes 50.5 and 88.9 is 0.00482.

Calculation:

The predicted value for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9 is calculated as follows:

$\begin{array}{c}\widehat{y}=0.5255-0.0236\left(-0.3747\right)+0.0097\left(-0.2769\right)\\ =0.5255+0.0088-0.0027\\ \approx 0.532\end{array}$

Thus, the predicted value for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9 is 0.532.

95% prediction interval:

The confidence interval is calculated using the formula:

$\widehat{Y_i}\pm t_{\frac{\alpha }{2},n-\left(k+1\right)}\sqrt{s^2+{\left(s_{{\widehat{\beta }}_i}\right)}^2}$

Where,

${\widehat{Y}}_i$ is the estimated value of the dependent variable.

$\alpha$ is the level of significance.

n is the total number of observations.

k is the total number of predictors in the model.

$s_{{\widehat{Y}}_i}$ is the standard error while calculating the estimated value of the dependent variable.

s is the overall standard deviation obtained after fitting the model.

Critical value:

Software procedure:

Step-by-step procedure to find the critical value is given below:

• Click on Graph, select View Probability and click OK.
• Select t, enter 17 as Degrees of freedom, in Shaded Area Tab select Probability under Define Shaded Area By and choose Both tails.
• Enter Probability value as 0.05.
• Click OK.

Output obtained from MINITAB is given below:

The 95% prediction interval is given below:

$\begin{array}{c}\widehat{Y_i}\pm t_{\frac{\alpha }{2},n-\left(k+1\right)}\sqrt{s^2{\left(s_{{\widehat{\beta }}_i}\right)}^2}=0.532\pm t_{\frac{0.05}{2},20-\left(2+1\right)}\sqrt{{\left(0.02001\right)}^2+{\left(0.00482\right)}^2}\\ =0.532\pm t_{0.025,17}\sqrt{\left(0.0004\right)+\left(0.000023\right)}\\ =0.532\pm \left(2.110\right)\left(0.0206\right)\\ =0.532\pm 0.043\end{array}$

$=0.489,0.575$

Thus, the 95% prediction interval for the specific gravity when the percentage of spring wood is 50.5 and percentage of light absorption in summerwood is 88.9 is (0.489,0.575).